Question

==Question solved==

This code will get you the dct2 output.

I am working to make improvement on my own codes

Thank you.

%Get the size of the input image
[m, n] = size(image);
output = zeros(m,n);

for u = 0:m-1
    for v = 0:n-1
        if u==0
            a=sqrt(1/8);
        else
            a=sqrt(2/8);
        end

        if v==0
            b=sqrt(1/8);
        else
            b=sqrt(2/8);
        end

        temp = 0;
        for x = 0:m-1
            for y = 0:n-1
                temp = (cos((((2*x)+1)*pi*u)/(2*m))*cos((((2*y)+1)*pi*v)/(2*n)));
            end
        end

        output(u+1,v+1) = a*b*temp;
    end
end

Answer 1

There are two problems with your code if you are trying to compare it to the Matlab function dct2. For the record, you can open and analyze the files to understand what Matlab is doing.

In both case you are not following the formula:

1. a and b need to be a function of n:

   if u==0
       a=sqrt(1/n);
   else
       a=sqrt(2/n);
   end
   
   if v==0
       b=sqrt(1/n);
   else
       b=sqrt(2/n);
   end
   

2. You need to multiply your temp term by the image values (which you are not doing):

   temp = temp+(image(x+1,y+1))*(cos((((2*x)+1)*pi*u)/(2*m))*cos((((2*y)+1)*pi*v)/(2*n)));
   

I haven’t addressed the performance problem, and you really should look into that on your own.